Finding the Kinetic Energy and the Speed of an Alpha particle

[boguslava]

Homework Statement

The nucleus of beryllium with mass number 8 decays into two alpha particles. The nuclear masses are M(8Be)=7456.6 MeV/c^2 and m(4He)=3727.4 MeV/c2. Assume that 8Be decays at rest and that we can neglect the atomic electrons. We are asked to find
a) the kinetic energy of each of the alpha particles
b) at what fraction of the speed of light is each outgoing alpha particle moving

Homework Equations

for a) I am using that the Kinetic energy equals the ratio of the alpha particle to the beryllium multiplied by the energy released as heat; KE=4/8xQ
for b) I am using the standard definition of KE=mv^2/2

The Attempt at a Solution

I am finding that the kinetic energy is 1864.6MeV for part a) and I am using this answer along with the mass of helium, which is 4u, to find the speed of the alpha particle. However, my calculation ends up showing that the alpha particle is moving with the speed of light. Is this realistic?

 

[Curious3141]

Homework Statement

The nucleus of beryllium with mass number 8 decays into two alpha particles. The nuclear masses are M(8Be)=7456.6 MeV/c^2 and m(4He)=3727.4 MeV/c2. Assume that 8Be decays at rest and that we can neglect the atomic electrons. We are asked to find
a) the kinetic energy of each of the alpha particles
b) at what fraction of the speed of light is each outgoing alpha particle moving

Homework Equations

for a) I am using that the Kinetic energy equals the ratio of the alpha particle to the beryllium multiplied by the energy released as heat; KE=4/8xQ

I don't understand where this equation comes from. What energy is being released as heat? How is it being released? Remember there's no electromagnetic radiation being released here, and no other matter to which the energy can be transferred, so how can heat be released?

What's simply happening is that the nucleus is undergoing fission into two alpha particles which travel in exactly opposite directions (as dictated by conservation of linear momentum). The only energy being imparted is kinetic energy of the alpha particles.

for b) I am using the standard definition of KE=mv^2/2

No. You should be using the Special Relativistic (SR) formulations for everything. Do you know what the SR equation for Kinetic Energy is?

Use Conservation of Energy, [itex]E = mc^2[/itex] and [itex]K = m_0c^2(\frac{1}{1 - {(\frac{v}{c})}^2} - 1)[/itex] here.

 

[boguslava]

Ok, I see what you are saying:
The kinetic energy of the alpha particle=Binding Energy of beryllium - Binding Energy of the two Helium atoms

 

[Curious3141]

Ok, I see what you are saying:
The kinetic energy of the alpha particle=Binding Energy of beryllium - Binding Energy of the two Helium atoms

Actually, the K.E. of each alpha particle is half the difference. 2 alpha particles are released after fission, and both have the same speed and therefore, energy.

Write the conservation statement this way:

[tex]Mc^2 = 2mc^2 + 2K[/tex]

where M is the mass of the berrylium atom, and m is the mass of one alpha particle. Don't convert the units, because all the (1/c^2) factors will happily cancel out with the c^2 terms. Work out K in eV.

 

[boguslava]

Thank you for this feedback! However, I am still confused as to how to show what fraction of the speed of light this kinetic energy for each helium atom is.
If I use the relativitity formula, do I use the mass of each helium atom separately, and what units do I use?

 

[Curious3141]

Thank you for this feedback! However, I am still confused as to how to show what fraction of the speed of light this kinetic energy for each helium atom is.
If I use the relativitity formula, do I use the mass of each helium atom separately, and what units do I use?

What answer did you get for the first part (easy)? Even though the answer (in MeV) is fairly compact, remember to respect the number of significant digits!

Once you have a value for K, do the algebra based on the equation I quoted in post #2. [itex]m_0[/itex] is simply the rest mass (as given) of a single alpha particle. Do all the algebraic rearrangement in symbols first, you're solving for [itex]\frac{v}{c}[/itex]. Then plug in the numbers. Keep the mass unit in terms of [itex]\frac{MeV}{c^2}[/itex] since the [itex]c^2[/itex] will cancel out in [itex]mc^2[/itex]. Again, remember that you need to express the answer to the appropriate number of significant figures.

 

[boguslava]

Thank you for the feedback! However, your formula for the relativistic kinetik energy is missing a root where the lorenz factor is defined. Upon calculations, I found that:
1) according to relativity v of alpha particle = 0.021971251c
2) according to classical physics v of alpha particle = 0.021975213c

This is a small difference, but still relativity is more precise.

 

[Curious3141]

Thank you for the feedback! However, your formula for the relativistic kinetik energy is missing a root where the lorenz factor is defined.

Oops. You're quite right, of course. I'd written it down correctly on paper when I was working on your question, but when I transcribed it to Latex, I missed the root. I hate Latex!

Should be:

[itex]K = m_0c^2(\frac{1}{\sqrt{1 - {(\frac{v}{c})}^2}} - 1)[/itex]

Upon calculations, I found that:
1) according to relativity v of alpha particle = 0.021971251c
2) according to classical physics v of alpha particle = 0.021975213c

This is a small difference, but still relativity is more precise.

The SR value is what I got as well. But remember your sig figs! Shouldn't you round to 5 sig figs?

And your answer to part a) should have the requisite number of trailing zeroes to make 5 sig figs too.